Question

Find the vector equation of a plane which is at a distance of 3 units from the origin and has $\hat{k}$ as the unit vector normal to it.

Answer 1

$$\begin{gathered} \text{Given that} \\ \text{normal vector,}\overrightarrow{n}\text{=}\hat{i} \\ \text{Now,}\hat{n}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{\hat{k}}{\left|\hat{k}\right|}=\frac{\hat{k}}{1}=\hat{k} \\ \text{The equation of a plane in normal form is} \\ \overrightarrow{r}.\hat{n}=d\text{(where}\mathit{\text{d}}\text{is the distance of the plane from the origin)} \\ \text{Substituting}\hat{n}\text{=}\hat{k}\text{and}\mathit{\text{d}}\text{= 3 in the relation, we get} \\ \overrightarrow{r}.\hat{k}=3 \end{gathered}$$